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%I #18 Feb 07 2020 13:39:11
%S 1,1,11,217,7691,430921,35117531,3927676537,577640740331,
%T 108115035641641,25097054302205051,7076531411753120857,
%U 2382432541064412524171,943997056642739165681161,434864796716131476530668571,230460477665217932140097413177
%N E.g.f.: Sum_{n>=0} exp(2*n*x) * Product_{k=1..n} (exp((2*k-1)*x) - 1).
%C Compare the e.g.f. to the identity:
%C exp(-x) = Sum_{n>=0} exp(2*n*x) * Product_{k=1..n} (1 - exp((2*k-1)*x)).
%H Vaclav Kotesovec, <a href="/A214687/b214687.txt">Table of n, a(n) for n = 0..175</a>
%H Hsien-Kuei Hwang, Emma Yu Jin, <a href="https://arxiv.org/abs/1911.06690">Asymptotics and statistics on Fishburn matrices and their generalizations</a>, arXiv:1911.06690 [math.CO], 2019.
%F E.g.f. A(x) satisfies: A(x) = exp(-x)*(2*G(x) - 1),
%F where G(x) = Sum_{n>=0} Product_{k=1..n} (exp((2*k-1)*x) - 1) = e.g.f. of A215066.
%F a(n) ~ 2*sqrt(6) * 24^n * (n!)^2 / (sqrt(n) * Pi^(2*n+3/2)). - _Vaclav Kotesovec_, May 05 2014
%e E.g.f.: A(x) = 1 + x + 11*x^2/2! + 217*x^3/3! + 7691*x^4/4! + 430921*x^5/5! +...
%e such that, by definition,
%e A(x) = 1 + exp(2*x)*(exp(x)-1) + exp(4*x)*(exp(x)-1)*(exp(3*x)-1)
%e + exp(6*x)*(exp(x)-1)*(exp(3*x)-1)*(exp(5*x)-1)
%e + exp(8*x)*(exp(x)-1)*(exp(3*x)-1)*(exp(5*x)-1)*(exp(7*x)-1) +...
%e Compare this series to the identity:
%e exp(-x) = 1 - exp(2*x)*(exp(x)-1) + exp(4*x)*(exp(x)-1)*(exp(3*x)-1)
%e - exp(6*x)*(exp(x)-1)*(exp(3*x)-1)*(exp(5*x)-1)
%e + exp(8*x)*(exp(x)-1)*(exp(3*x)-1)*(exp(5*x)-1)*(exp(7*x)-1) +-...
%e The related e.g.f. of A215066 equals the series:
%e G(x) = 1 + (exp(x)-1) + (exp(x)-1)*(exp(3*x)-1)
%e + (exp(x)-1)*(exp(3*x)-1)*(exp(5*x)-1)
%e + (exp(x)-1)*(exp(3*x)-1)*(exp(5*x)-1)*(exp(7*x)-1) +...
%e or, more explicitly,
%e G(x) = 1 + x + 7*x^2/2! + 127*x^3/3! + 4315*x^4/4! + 235831*x^5/5! +...
%e such that G(x) satisfies:
%e G(x) = (1 + exp(x)*A(x))/2.
%o (PARI) {a(n)=n!*polcoeff(sum(m=0,n+1,exp(2*m*x+x*O(x^n))*prod(k=1,m,exp((2*k-1)*x+x*O(x^n))-1)),n)}
%o for(n=0,26,print1(a(n),", "))
%Y Cf. A207214, A215066.
%K nonn
%O 0,3
%A _Paul D. Hanna_, Aug 01 2012
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